ICSE Class 10 Computer Chapter 2 VSA Solutions: Java

Question 48: Write the Java statement for the sum of a raised to b and the cube root of c.

Answer:

double ans = Math.pow(a, b) + Math.cbrt(c);

Working: The phrase “a raised to b” is written as Math.pow(a, b). The cube root of c is written as Math.cbrt(c). Since both methods return a double, the suitable variable type is double.

If the question asks only for the expression, the required expression is:

Math.pow(a, b) + Math.cbrt(c)

Question 49: Name the following.

(a) Method with the same name as the class and invoked every time an object is created.

Answer: Constructor.

Reason: A constructor has the same name as its class and is called automatically when an object of that class is created.

(b) Keyword used to access the classes of a package.

Answer: import.

Reason: The import keyword allows a program to use classes from a package, such as java.util.Scanner.

Question 50: Name the following Java methods.

(a) A Character method that checks whether a character is an alphabet or a number.

Answer: Character.isLetterOrDigit(ch).

Reason: This method returns true when ch is either a letter or a digit.

(b) A Math method that does not have an argument.

Answer: Math.random().

Reason: Math.random() does not need an input argument. It returns a random double value in the range from 0.0 up to, but not including, 1.0.

Question 52: Fill in the blanks to count names ending with KUMAR.

void count(String s[])
{
    int i, l=_________, c=0;
    for(i=0;i<l;i++)
    {
        if(________________)
            c++;
    }
    System.out.println(c);
}

Answer:

void count(String s[])
{
    int i, l = s.length, c = 0;
    for(i = 0; i < l; i++)
    {
        if(s[i].endsWith("KUMAR"))
            c++;
    }
    System.out.println(c);
}

Working: The number of elements in the array is obtained by s.length. Each name is one element of the array, so the current name is s[i]. The method endsWith("KUMAR") checks whether that name ends with the exact suffix KUMAR.

Important note: endsWith() is case-sensitive. The string "Ravi Kumar" does not end with "KUMAR" unless it is converted or stored in uppercase.

Question 53: Extract PASS and PASSION from COMPASSION.

Answer:

String word = "COMPASSION";
System.out.println(word.substring(3, 7));  // PASS
System.out.println(word.substring(3));     // PASSION

Working: Java string indexing starts from 0.

For PASS, the substring starts at index 3 and must include the character at index 6. Since the ending index in substring(start, end) is excluded, we use 7 as the ending index.

For PASSION, the substring starts at index 3 and continues to the end, so substring(3) is enough.

Question 55: Correct the error in the character array program.

char arr[]={'4', '&', 'a', 'w', 'd'};
int i;
for(i=0;i<arr.length;i++)
{
    if(Character.isLetter(arr))
        System.out.println(arr);
}

Corrected code:

char arr[] = {'4', '&', 'a', 'w', 'd'};
int i;
for(i = 0; i < arr.length; i++)
{
    if(Character.isLetter(arr[i]))
        System.out.println(arr[i]);
}

Explanation: The method Character.isLetter() checks one character at a time. The name arr refers to the whole array, not one element. Inside the loop, the current character is arr[i], so both the condition and the print statement must use arr[i].

Output of the corrected program:

a
w
d

Question 56: Name the type of type casting.

(a)

double m = 'b';

Answer: Implicit type casting, also called widening conversion.

Reason: A char value can be converted automatically to a double. Java uses the character’s Unicode value in the conversion. The character 'b' has the value 98, so m receives 98.0.

(b)

char ch = (char)68;

Answer: Explicit type casting, also called narrowing conversion.

Reason: The cast operator (char) tells Java to convert the integer value 68 into its corresponding character. In Unicode/ASCII for English uppercase letters, 68 represents 'D'.

Question 57: Write Java statements using suitable data types.

(a) Assign the cube root of -343 to a variable.

Answer:

double root = Math.cbrt(-343);

Working: The method Math.cbrt() returns a double. The cube root of -343 is -7.0, because (-7) * (-7) * (-7) = -343.

(b) Assign the position of the last occurrence of @ in the string s.

Answer:

int pos = s.lastIndexOf('@');

Working: A string index is an integer position. Therefore, the suitable data type is int. The method lastIndexOf('@') returns the last position of @, or -1 if it is absent.

Question 58: Mention the output of the code snippet.

int lives = 5;
System.out.println(lives--);
System.out.println(lives);

Answer:

5
4

Working: The expression lives-- uses the postfix decrement operator. In postfix form, Java first uses the current value and then decreases it by 1. Therefore, the first print statement prints 5. After that, lives becomes 4, so the second print statement prints 4.

Question 59: Convert the for loop to an exit-controlled loop.

for(k = 10; k >= -1; k--)
    System.out.println(k*2);
System.out.println(k*4);

Answer:

k = 10;
do
{
    System.out.println(k * 2);
    k--;
} while(k >= -1);

System.out.println(k * 4);

Working: An exit-controlled loop checks its condition after running the loop body, so the correct form is do-while. The initialization k = 10 must come before the loop. The update k-- must be placed inside the loop body after the print statement.

Key point: In the original code, there are no braces after the for statement. Therefore, only System.out.println(k*2); belongs to the loop. The statement System.out.println(k*4); runs once after the loop finishes, so it must remain outside the do-while loop.